Chapter 12

Static Equilibrium And Elasticity

(Keseimbangan Statik dan Kekenyalan)

Static Equilibrium Equilibrium implies the object is at

rest (static) or its center of mass moves with a constant velocity (dynamic)

Static equilibrium is a common situation in engineering

Principles involved are of particular interest to civil engineers, architects, and mechanical engineers

Conditions for Equilibrium The net force equals zero

F = 0 If the object is modeled as a particle,

then this is the only condition that must be satisfied

The net torque equals zero = 0 This is needed if the object cannot be

modeled as a particle

Torque = r x F

Use the right-hand rule to determine the direction of the torque

The tendency of the force to cause a rotation about O depends on F and the moment arm d

Rotational Equilibrium Need the angular acceleration of

the object to be zero For rotation, = I

For rotational equilibrium, = 0 This must be true for any axis of

rotation

Equilibrium Summary There are two necessary conditions for

equilibrium The resultant external force must equal zero:

F = 0 This is a statement of translational equilibrium The acceleration of the center of mass of the object

must be zero when viewed from an inertial frame of reference

The resultant external torque about any axis must be zero: = 0

This is a statement of rotational equilibrium The angular acceleration must equal zero

Static vs. Dynamic Equilibrium In this chapter, we will concentrate

on static equilibrium The object will not be moving

Dynamic equilibrium is also possible The object would be rotating with a

constant angular velocity In either case, the = 0

Equilibrium Equations We will restrict the applications to

situations in which all the forces lie in the xy plane These are called coplanar forces since

they lie in the same plane There are three resulting equations

Fx = 0 Fy = 0 z = 0

Axis of Rotation for Torque Equation The net torque is about an axis

through any point in the xy plane The choice of an axis is arbitrary If an object is in translational

equilibrium and the net torque is zero about one axis, then the net torque must be zero about any other axis

Center of Gravity All the various

gravitational forces acting on all the various mass elements are equivalent to a single gravitational force acting through a single point called the center of gravity (CG)

Center of Gravity, cont The torque due to the gravitational force

on an object of mass M is the force Mg acting at the center of gravity of the object

If g is uniform over the object, then the center of gravity of the object coincides with its center of mass

If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center

Problem-Solving Strategy – Equilibrium Problems Draw a diagram of the system Isolate the object being analyzed

Draw a free-body diagram Show and label all external forces acting

on the object Indicate the locations of all the forces For systems with multiple objects, draw a

separate free-body diagram for each object

Problem-Solving Strategy – Equilibrium Problems, 2 Establish a convenient coordinate

system Find the components of the forces

along the two axes Apply the first condition for

equilibrium (F=0) Be careful of signs

Problem-Solving Strategy – Equilibrium Problems, 3 Choose a convenient axis for

calculating the net torque on the object Remember the choice of the axis is

arbitrary Choose an origin that simplifies the

calculations as much as possible A force that acts along a line passing

through the origin produces a zero torque

Problem-Solving Strategy – Equilibrium Problems, 4 Apply the second condition for

equilibrium ( = 0) The two conditions of equilibrium

will give a system of equations Solve the equations simultaneously If the solution gives a negative for a

force, it is in the opposite direction than what you drew in the free body diagram

Weighted Hand Example Model the forearm

as a rigid bar The weight of the

forearm is ignored There are no forces

in the x-direction Apply the first

condition for equilibrium (Fy = 0)

Weighted Hand Example, cont Apply the second

condition for equilibrium using the joint O as the axis of rotation ( =0)

Generate the equilibrium conditions from the free-body diagram

Solve for the unknown forces (F and R)

Horizontal Beam Example The beam is uniform

So the center of gravity is at the geometric center of the beam

The person is standing on the beam

What are the tension in the cable and the force exerted by the wall on the beam?

Horizontal Beam Example, 2 Draw a free-body

diagram Use the pivot in the

problem (at the wall) as the pivot This will generally be

easiest Note there are

three unknowns (T, R, )

The forces can be resolved into components in the free-body diagram

Apply the two conditions of equilibrium to obtain three equations

Solve for the unknowns

Horizontal Beam Example, 3

Ladder Example The ladder is

uniform So the weight of the

ladder acts through its geometric center (its center of gravity)

There is static friction between the ladder and the ground

Ladder Example, 2 Draw a free-body

diagram for the ladder The frictional force is ƒ

= µn Let O be the axis of

rotation Apply the equations

for the two conditions of equilibrium

Solve the equations

Add a person of mass M at a distance d from the base of the ladder

The higher the person climbs, the larger the angle at the base needs to be

Eventually, the ladder may slip

Ladder Example, Extended

Elasticity So far we have assumed that

objects remain rigid when external forces act on them Except springs

Actually, objects are deformable It is possible to change the size

and/or shape of the object by applying external forces

Internal forces resist the deformation

Definitions Associated With Deformation Stress

Is proportional to the force causing the deformation

It is the external force acting on the object per unit area

Strain Is the result of a stress Is a measure of the degree of

deformation

Elastic Modulus The elastic modulus is the constant of

proportionality between the stress and the strain For sufficiently small stresses, the stress

is directly proportional to the stress It depends on the material being

deformed It also depends on the nature of the

deformation

Elastic Modulus, cont The elastic modulus in general

relates what is done to a solid object to how that object responds

Various types of deformation have unique elastic moduli

strainstress

ulusmodelastic

Three Types of Moduli Young’s Modulus

Measures the resistance of a solid to a change in its length

Shear Modulus Measures the resistance to motion of

the planes within a solid parallel to each other

Bulk Modulus Measures the resistance of solids or

liquids to changes in their volume

Young’s Modulus The bar is stretched

by an amount L under the action of the force F

The tensile stress is the ratio of the external force to the cross-sectional area A

Young’s Modulus, cont The tension strain is the ratio of the

change in length to the original length

Young’s modulus, Y, is the ratio of those two ratios:

Units are N / m2

tensile stress

tensile straini

FAYLL

Stress vs. Strain Curve Experiments show

that for certain stresses, the stress is directly proportional to the strain This is the elastic

behavior part of the curve

Stress vs. Strain Curve, cont The elastic limit is the maximum

stress that can be applied to the substance before it becomes permanently deformed

When the stress exceeds the elastic limit, the substance will be permanently defined The curve is no longer a straight line

With additional stress, the material ultimately breaks

Shear Modulus Another type of

deformation occurs when a force acts parallel to one of its faces while the opposite face is held fixed by another force

This is called a shear stress

Shear Modulus, cont For small deformations, no change in

volume occurs with this deformation A good first approximation

The shear stress is F / A F is the tangential force A is the area of the face being sheared

The shear strain is x / h x is the horizontal distance the sheared

face moves h is the height of the object

Shear Modulus, final The shear modulus is the ratio of

the shear stress to the shear strain

Units are N / m2

shear stress

shear strain

FASxh

Bulk Modulus Another type of

deformation occurs when a force of uniform magnitude is applied perpendicularly over the entire surface of the object

The object will undergo a change in volume, but not in shape

Bulk Modulus, cont The volume stress is defined as the

ratio of the magnitude of the total force, F, exerted on the surface to the area, A, of the surface

This is also called the pressure The volume strain is the ratio of

the change in volume to the original volume

Bulk Modulus, final The bulk modulus is the ratio of the

volume stress to the volume strain

The negative indicates that an increase in pressure will result in a decrease in volume

volume stress

volume straini i

F PABV VV V

Compressibility The compressibility is the inverse

of the bulk modulus It often used instead of the bulk

modulus

Moduli and Types of Materials Both solids and liquids have a bulk

modulus Liquids cannot sustain a shearing

stress or a tensile stress If a shearing force or a tensile force is

applied to a liquid, the liquid will flow in response

Moduli Values

Prestressed Concrete

If the stress on a solid object exceeds a certain value, the object fractures

The slab can be strengthened by the use of steel rods to reinforce the concrete

The concrete is stronger under compression than under tension

Prestressed Concrete, cont A significant increase in shear strength is

achieved if the reinforced concrete is prestressed

As the concrete is being poured, the steel rods are held under tension by external forces

These external forces are released after the concrete cures

This results in a permanent tension in the steel and hence a compressive stress on the concrete

This permits the concrete to support a much heavier load